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For the functional $J$ defined by $$J(y)=\int_0^1y'\sqrt{1+(y'')^2}dx,$$Find an extremal satisfying the conditions $y(0)=0,~y'(0)=0,~y(1)=1$ and $y'(1)=2$?

My attempt:

Let $F(x,y,y',y'')=y'\sqrt{1+(y'')^2}$, then second order Euler's equation $$F_y-\frac{d}{dx}F_{y'}+\frac{d^2}{dx^2}F_{y''}=0,$$ gives $$0-\frac{d}{dx}\sqrt{1+(y'')^2}+\frac{d^2}{dx^2}\left(\frac{y' \cdot y''}{\sqrt{1+(y'')^2}}\right)=0,$$ How to proceed now with simple steps? It seem very tedious to solve the equation.

Messi Lio
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    If I remembered correctly for this kind of functional without time variable $x$ you can try the Beltrami identity to avoid these second order differential equations – Richard Chen Apr 27 '22 at 07:47
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    This is basically integrating once the Euler-Lagrange equation (which is easy as it is in the form $d/dx(\ldots) = 0$. Equivalently, this is rephrasing the problem in terms of $z= y'$. The integration constant is the same as the Lagrange multiplier we need to enforce the constraint $\int_0^1 z\text dx = 1$. – SolubleFish Apr 27 '22 at 09:20

1 Answers1

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As I point out in the comment, we can rephrase the problem in terms of $z= y'$. The initial conditions $y(0) = 0$, $y(1) = 1$ then became the constraint $\int_0^1 z\text dx = 1$. OP's variational problem is therefore equivalent to the following one : $$\tilde J[z] = \int_0^1 (z\sqrt{1+z'^2} - Cz)\text dx $$ where $C$ is a Lagrange multiplier and with boundary conditions $z(1)=2$, $z(0) = 0$

In other words, we have the Lagrangian $L(z,z') = z\sqrt{1+z'^2} - Cz$.

Since this is independent of $x$, we know that : $$H = pz' - L$$ is conserved along the solutions of the Euler-Lagrange equation, where $p = \frac{\partial L}{\partial z'}$. Explicitly, we have : \begin{align} p &= \frac{zz'}{\sqrt{1+z'^2}} \\ H &= Cz - \frac{z}{\sqrt{1+z'^2}} = E \tag 1 \end{align} for some constant $E$.

This is then rewritten : $$1+z'^2 = \frac{z^2}{(Cz-E)^2} \tag 2$$

Edit @Diger provided the end of the solution in the comments below. I reproduced them here to provide a complete answer.

Evaluating $(1)$ at $x= 0$ and using $z(0) = 0$, we get $E= 0$. Then, $(2)$ becomes : $$1+z'^2 = \frac{1}{C^2}$$

ie $z'$ is constant and $z(x) = ax+b$. The boundary conditions then fix $a=2, b=0$. Then $\int_0^1 z(x) \text dx$ is automatically satisfied.

Therefore, $y(x) = \int_0^x z(x')\text dx' = x^2$ is the solution.

SolubleFish
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  • I don't think $\int_0^1 z {\rm d}x =1$ is equivalent to $y(0)=0$ and $y(1)=1$. I guess you still need to enforce $y(0)=0$ at the end. – Diger Apr 27 '22 at 10:52
  • The problem is only solvable, because the boundary conditions are "lucky". First you can see from the equation, by letting $x\rightarrow 0$, that $E=0$. The equation then becomes $$z'^2+1=1/C^2$$ or $$z=cx+c_1, ,$$ for constants $c,c_1$. Boundary conditions $z(0)=0$ and $z(1)=2$ give $$z=2x,,$$ which automatically satisfies $$\int_0^1 {\rm d}x , 2x =1,.$$ Then $$y(x)=\int_0^x {\rm d}t , 2t = x^2 , .$$ – Diger Apr 27 '22 at 11:20
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    @Diger Since the EL equation only involves derivatives of $y$, you can shift it by a constant. Therefore the only relevant boundary condition is $y(1) -y(0) = 1$. This is what I meant above. – SolubleFish Apr 27 '22 at 11:54